∵OD为半径
∴AB是⊙O的切线 ························································································· 5分 （2）连接OF，令半径＝r，则OF＝r，CF＝4，CO＝r－3
利用勾股定理可得，r＝6 ……………………………………………………………………7分
∵EF∥AB，∴∠CEF＝∠B． ∴△CEF∽△DBO ∴
CFCE
＝ DO DB
25
25
∴BD＝8 ····································································································· 9分 27．（本题10分） 解：
（1）① 2 cm ······························································································· 2分 ② 分点C、P在BQ同侧和异侧两种情况，画对一种就给全分
B
QB
C O P
QC
O P A
图2
图1
A
·················································································································· 5分 （2）当点C在∠AOB的内部或一边上时，则重叠部分即为△CPQ.
因为△CPQ是由△OPQ折叠得到，所以当△OPQ为等腰三角形时，重叠部分必为等腰三角形.如图1、2、3三种情况： BBB Q
Q Q
O O O A A PP P
图2 图3 图1
九年级数学 第 11 页 （共 12 页）
A
当QO＝QP时， 当PQ＝PO时， 当OQ＝OP时， OQ＝错误！未找到引用
OQ＝错误！未找到引用源。OQ＝OP＝2cm 源。OP＝2cm 当点C在∠AOB的外部时，
B B Q
C
Q
O O P P A A
图4 图5
C
当点C在射线OB的上方时（如图4）， 当点C在射线OA的下方时（如图5），
OQ＝6－2(cm) OQ＝6＋2(cm)
共5种情况，每画对1种情况得1分，共5分.
················································································································· 10分
九年级数学 第 12 页 （共 12 页）